Question

Newton's law of cooling states that the temperature of an object changes at a net proportional to the difference between its temperature ad that of its surroundings. Suppose that the temperature of a cup of coffee obeys hew of cooling If the coffee has a temperature of 200 . F when freshly poured, and 1 min later has cooled to \(190^{-} \mathrm{F}\) in at \(70^{\circ} \mathrm{F}\), determine when the coffer reaches a temperature of \(150^{\circ} \mathrm{F}\).

Short answer

Answer: It takes approximately \(33.8\) minutes for the coffee to cool down to \(150^\circ\mathrm{F}\) in a \(70^\circ\mathrm{F}\) room.

Step-by-step solution

Calculate the constant (k) using initial temperature and temperature after 1 minute

Let's use the initial temperature and the temperature after 1 minute to calculate the value of \(k\). We know that the initial temperature is \(200^\circ\mathrm{F}\) and the temperature after 1 minute is \(190^\circ\mathrm{F}\). The surrounding temperature is \(70^\circ\mathrm{F}\). We'll plug these values into the formula: \(\frac{dT}{dt} = k(T - T_s)\) \(\frac{190 - 200}{1} = k(200 - 70)\) Solving for \(k\), we get: \(k = \frac{-10}{130} = -\frac{1}{13}\)

Use k to find the time it takes to cool down to \(150^\circ\mathrm{F}\)

Now that we have the value of \(k\), we'll use it to find the time it takes for the coffee to cool down to \(150^\circ\mathrm{F}\). We'll plug the value of \(k\) and the final temperature of \(150^\circ\mathrm{F}\) into the formula: \(\frac{dT}{dt} = k(T - T_s)\) \(\frac{150 - T}{t} = -\frac{1}{13}(T - 70)\) Now, we'll solve for \(t\): \((150 - T)t = -\frac{13}{1}(T - 70)\) Plug the initial temperature (\(200^\circ\mathrm{F}\)): \((150 - 200)t = -\frac{13}{1}(200 - 70)\) Solving this equation for \(t\), we get: \(t = \frac{-13(200 - 70)}{-50} = \frac{1690}{50} = 33.8\) minutes

Present the final answer

Therefore, it takes approximately \(33.8\) minutes for the coffee to cool down to \(150^\circ\mathrm{F}\) in a \(70^\circ\mathrm{F}\) room.

Key concepts

Differential Equations

Understanding differential equations is crucial to solving many problems in physics and engineering, including the process of cooling described by Newton's law. A differential equation is a mathematical equation that relates some function with its derivatives. When dealing with physical processes such as temperature change, these equations describe how a rate of change (in this case, temperature change rate) is related to the current state of a system.

For Newton's law of cooling, the differential equation takes the form \(\frac{dT}{dt} = k(T - T_s)\), where \(\frac{dT}{dt}\) represents the rate of temperature change, \(T\) is the temperature of the object, \(T_s\) is the surrounding temperature, and \(k\) is a constant that represents the cooling rate. Solving a differential equation involves finding the function \(T(t)\) that satisfies the given relationship for all values of \(t\), which is the time variable. The solution often requires initial conditions to be set, such as the initial temperature of the coffee.

Exponential Decay

The concept of exponential decay is a cornerstone in understanding processes like cooling, radioactive decay, and population decrease. Exponential decay describes a process where the quantity decreases at a rate proportional to its current value. In terms of the cooling problem, as the coffee cools, the rate at which it loses heat to its surroundings is proportional to the difference between its temperature and the room temperature.

Mathematically, this is expressed by the same differential equation used in Newton's law of cooling. When solved, it leads to an exponential function that models the temperature of the coffee over time, \(T(t) = T_s + (T_0 - T_s)e^{-kt}\), where \(T_0\) is the initial temperature of the coffee, and \(e\) is the base of the natural logarithm. This model shows that as time increases, the temperature difference between the coffee and the surroundings decreases exponentially, indicating that the coffee cools down rapidly at first, but the rate slows down as it approaches the surrounding temperature.

Temperature Change Rate

The temperature change rate is an essential aspect of Newton's law of cooling and can be understood as how quickly an object's temperature approaches the temperature of its surroundings. This rate depends on several factors, including the material properties, the geometry of the object, and the difference in temperature between the object and its environment

In our exercise, the rate at which the coffee cools is given by the derivative \(\frac{dT}{dt}\). By solving the differential equation, we determine the constant \(k\), which reflects how effectively the coffee transfers heat to the air. A higher absolute value of \(k\) would indicate faster cooling. The temperature change rate allows us to predict not just when the coffee will reach a particular temperature like \(150^\circ\mathrm{F}\), but also to describe its cooling behavior at every moment in time. This has practical applications, for example, in food safety, where cooling rates are essential to prevent bacterial growth.